of a slab of thickness t and with thermal conductivity k is The temperature gradient, given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=01, 0()=0 2 and Fourier's law Q = -kA d0 show that (02-0,) Q = -kA- [6(0)=0, means that when x is 0 then 0301, and 0(f)=02 means that when x is t then 0(t) =02.)

of a slab of thickness t and with thermal conductivity k is The temperature gradient, given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=01, 0()=0 2 and Fourier's law Q = -kA d0 show that (02-0,) Q = -kA- [6(0)=0, means that when x is 0 then 0301, and 0(f)=02 means that when x is t then 0(t) =02.)

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Differential Equations

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The temperature gradient,
, of a slab of thickness t and with thermal conductivity k is
given by
de
= C
dx
where C is a constant and 0 is a function of x.
By using the conditions 0(0)=0+, 0(1)=0 2 and Fourier's law
Q = -kA d0
dx
show that
(02-0,)
Q = -kA-
[(0)=0, means that when x is 0 then 0=01, and 0(t)=02 means that when x is t then 0(t) =02.]

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